= Solution
Let all <velocity> amplitudes refer to the same positive normal direction, from medium 1 into medium 2. For a right-going linear <sound wave>, its <pressure> amplitude is $p'=Zu'$ with <acoustic impedance> $Z=\rho c_s$; for a left-going wave it is $p'=-Zu'$. Continuity of normal <velocity> and <pressure> at the interface gives
$$
u_i+u_r=u_t,\qquad Z_1(u_i-u_r)=Z_2u_t.
$$
Eliminating $u_t$ yields the <normal-incidence acoustic velocity reflection coefficient>
$$
\frac{u_r}{u_i}=\frac{Z_1-Z_2}{Z_1+Z_2}.
$$
The common <adiabatic index> and mean particle mass $\bar m$ give $c_{s,j}=\sqrt{\gamma k_BT_j/\bar m}$, so
$$
\boxed{\frac{u_r}{u_i}=\frac{\rho_1\sqrt{T_1}-\rho_2\sqrt{T_2}}
{\rho_1\sqrt{T_1}+\rho_2\sqrt{T_2}}.}
$$
A negative coefficient reverses the reflected <velocity> perturbation's phase in this fixed coordinate system. The reflected pressure coefficient has the opposite sign. Pressure balance also gives $\rho_1T_1=\rho_2T_2$, so the impedance ratio can equivalently be written $Z_2/Z_1=\sqrt{\rho_2/\rho_1}$. These are normal-incidence, small-amplitude results without a surface mass or tension.
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